Question

Write a balanced equation for the oxidation of \(p\)-xylene to 1,4-benzenedicarboxylic acid (terephthalic acid) using potassium dichromate in aqueous sulfuric acid. How many milligrams of \(\mathrm{H}_{2} \mathrm{CrO}_{4}\) are required to oxidize \(250 \mathrm{mg}\) of \(p\) xylene to terephthalic acid?

Short answer

Answer: 1667 milligrams of H2CrO4 are required to oxidize 250 milligrams of p-xylene to terephthalic acid.

Step-by-step solution

Write the balanced equation for the reaction

First, we write the balanced equation: \(p\)-xylene + \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) + \(\mathrm{H}_{2}\mathrm{SO}_{4}\) \(\rightarrow\) terephthalic acid + \(\mathrm{K}_{2}\mathrm{SO}_{4}\) + \(\mathrm{Cr}_{2}\mathrm{SO}_{4}_3\) + H2O To balance this equation, we need to have equal amounts of each element on both sides of the equation. The balanced equation is: \[\mathrm{C}_{6}\mathrm{H}_{4}(CH_{3})_{2} + 6 \mathrm{H}_{2}\mathrm{CrO}_{4} \rightarrow \mathrm{C}_{6}\mathrm{H}_{4}(\mathrm{COOH})_{2} + 6\mathrm{H}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 4\mathrm{H}_{2}O\] This is the balanced equation for the oxidation of p-xylene to terephthalic acid using potassium dichromate in aqueous sulfuric acid.

Calculate the stoichiometry of the reaction

Now that we have the balanced equation, we can determine the stoichiometry of the reaction. From the equation, we can see that one mole of p-xylene reacts with six moles of \(\mathrm{H}_{2}\mathrm{CrO}_{4}\). To calculate the amount of \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) required to oxidize 250 mg of p-xylene, we need to convert the mass of p-xylene to moles using its molar mass. Molar mass of p-xylene = 106.16 g/mol Moles of p-xylene = (250 mg)/(106.16 g/mol) * (1 g/1000 mg) = 0.002354 mol p-xylene

Calculate the amount of H2CrO4 required for the oxidation

Now, using the stoichiometry of the reaction, we can calculate the moles of \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) required to oxidize the amount of p-xylene: Moles of \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) required = (0.002354 mol p-xylene) * (6 moles \(\mathrm{H}_{2}\mathrm{CrO}_{4}\)/1 mole p-xylene) = 0.01412 mol \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) Finally, to find the mass of \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) in milligrams, we need to multiply the moles of \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) by its molar mass and convert back to milligrams: Molar mass of \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) = 118.03 g/mol Mass of \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) required = (0.01412 mol \(\mathrm{H}_{2}\mathrm{CrO}_{4}\)) * (118.03 g/mol) * (1000 mg/g) = 1667 mg \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) Hence, 1667 milligrams of \(\mathrm{H}_{2}\mathrm{CrO}_{4}\) are required to oxidize 250 milligrams of p-xylene to terephthalic acid.

Key concepts

Balancing Chemical Equations

Understanding how to balance chemical equations is a fundamental skill in chemistry. When a chemical reaction occurs, atoms are neither created nor destroyed.

Due to this conservation of mass, the quantity of each element must remain the same before and after the reaction. Balancing a chemical equation ensures that the number of atoms for each element is the same on both the reactant and product sides of the equation. In the exercise, the oxidation of p-xylene to terephthalic acid, we see that six moles of dichromate ions are needed to oxidize one mole of p-xylene, resulting in equal numbers of carbon, hydrogen, and chromium atoms on each side of the equation.

To achieve this balance, you may need to add coefficients—whole numbers placed in front of the chemical formulas—to adjust the number of molecules or formula units of the reactants and products. This process can sometimes be trial and error, but there are systematic approaches, such as starting with the most complex molecule and balancing elements that appear in only one reactant and one product first.

Stoichiometry

Stoichiometry is the quantitative relationship between the reactants and products in a balanced chemical equation. This concept enables chemists to calculate the amounts of reactants needed or products formed in a chemical reaction.

In stoichiometry, the coefficients in the balanced equation represent the ratio of moles of each substance involved. In the given exercise, the ratio is 1 mole of p-xylene to 6 moles of dichromate ions. To find out how much of a reactant is needed to completely react with a given amount of another substance, you first convert the mass of the substance to moles and then use the molar ratio from the balanced equation to calculate the moles of the desired reactant or product.

Remember that the molar ratio is the heart of stoichiometry—it tells us, in this case, that it takes six times as much dichromate in moles to fully oxidize the p-xylene present.

Molar Mass Calculations

Molar mass calculations are vital when converting between mass and moles in chemistry. The molar mass, usually expressed in grams per mole (g/mol), is the mass of one mole of a substance. It numerically equals the average atomic or molecular weight of the substance in unified atomic mass units (u).

To solve problems in stoichiometry, one often needs to convert mass to moles or vice versa, and this is done using the molar mass as a conversion factor. The exercise showcases this conversion where the mass of p-xylene in milligrams is converted to moles using its molar mass. Then, using the stoichiometric ratio, the moles of dichromate required are calculated.

Finally, to find the mass of dichromate needed, the molar mass of dichromate is used to convert moles back to milligrams. It's essential to work with a consistent unit system when performing these calculations, often requiring conversion between grams and milligrams, for accurate results.